Problem 2 - The Calculus with Analytic Geometry

Question

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=x^{2}-6 x+9 $$

Step-by-Step Solution

Answer

The slope of the tangent line at \((x_{1}, y_{1})\) is \(2x_{1} - 6\). The graph has a horizontal tangent at (3, 0).

Step 1 - Find the Derivative of the Function

To find the slope of the tangent line to the graph of the function at any point, start by differentiating the function with respect to x. Given function: \[ y = x^2 - 6x + 9 \]The derivative is: \[ \frac{dy}{dx} = 2x - 6 \]

Step 2 - Evaluate the Derivative at the Point

To find the slope of the tangent at the given point \((x_{1}, y_{1})\), substitute the x-coordinate of the point into the derivative:\[ m = 2x_{1} - 6 \]

Step 3 - Find Points Where the Graph Has a Horizontal Tangent

A horizontal tangent occurs where the derivative is zero:\[ 2x - 6 = 0 \]Solving for x gives:\[ x = 3 \]Substitute x = 3 back into the original function to find the y-coordinate:\[ y = 3^2 - 6(3) + 9 = 0 \]Thus, the point where the graph has a horizontal tangent is (3, 0).

Step 4 - Create a Table of Values

Select a range of x values to calculate corresponding y values and slopes (m). For example:\[ \begin{array}{|c|c|c|} \hline x & y & m \ \hline 0 & 9 & -6 \ 1 & 4 & -4 \ 2 & 1 & -2 \ 3 & 0 & 0 \ 4 & 1 & 2 \ 5 & 4 & 4 \ 6 & 9 & 6 \ \hline \end{array} \]Include the points where the graph has a horizontal tangent.

Step 5 - Draw a Sketch of the Graph

Plot the points from the table and draw the parabola. Label the tangent lines and the point (3, 0) where the tangent is horizontal.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies continuous change. It has two primary areas: differential calculus and integral calculus. In this exercise, we focus on differential calculus, which involves finding the derivative of a function. The derivative represents the rate of change or the slope of a function at a given point. Understanding calculus is crucial for solving problems related to motion, growth, and many other real-world phenomena.

Derivative

A derivative is a mathematical tool that measures how a function changes as its input changes. It is denoted as \(\frac{dy}{dx}\) for a function \(y = f(x)\). To find the derivative of the given function \(y = x^2 - 6x + 9\), we use basic differentiation rules.
The derivative of \(x^2\) is \(2x\), the derivative of \(6x\) is \(6\), and the constant 9 disappears.
So, the derivative of our function is:
\[ \frac{dy}{dx} = 2x - 6 \]
This derivative, \(2x - 6\), gives us the slope of the tangent line to any point on the graph of the function.

Horizontal Tangent

A horizontal tangent line occurs when the slope of the tangent line is zero. This means the derivative of the function at that particular point is zero.
To find where this happens in our function \(y = x^2 - 6x + 9\), we set the derivative equal to zero:
\[ 2x - 6 = 0 \]
Solving this equation, we get \(x = 3\).
We then substitute \(x = 3\) back into the original function to find the corresponding \(y\) value:
\[ y = 3^2 - 6(3) + 9 = 0 \]
Thus, the point \((3, 0)\) is where the graph has a horizontal tangent.

Function Evaluation

Function evaluation involves calculating the value of a function for specific inputs. Using our function \(y = x^2 - 6x + 9\), we can create a table of values to see how \(y\) and the slope \(m\) change as \(x\) varies.
For example, at \(x = 0\), \(y = 9\), and the slope \((m)\) is \(-6\) because \[ m = 2(0) - 6 = -6 \]
Similarly, for \(x = 1\), we calculate \(y = 1^2 - 6(1) + 9 = 4\)
and the slope is:
\[ m = 2(1) - 6 = -4 \]
Repeat this for other values to get a full range.
This table provides a clear understanding of how the function behaves and where the slope changes, especially at the important point \((3, 0)\) where the tangent is horizontal.